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To: Zeuspaul who wrote (3743)	11/26/1998 4:17:00 PM
From: AreWeThereYet	Respond to of 14778

I believe VA = Watt * 1.4. I don't have time to dig up the textbook to verify but I am 80% certain about that. aC

To: Zeuspaul who wrote (3743)	11/26/1998 7:42:00 PM
From: Howard R. Hansen	Respond to of 14778

I thought Watts = Volts x Amps ( W=VA ) ie 600 VA = 600 watts. Am I misunderstanding something? For sinusoidial signals Watts = Volts x Amps x Cos( theta ). Where theta is the phase shift between the voltage and current wave forms. However, the current drawn by a computer is rarely a sinusoidial signal and in this case the calculation of watts is a lot more complicated. But in any case the power used by a computer will be significantly less than what is given by a multiplying volts times amps and the rating of the UPS should be based on VA, (Volts x Amps) not watts.

To: Zeuspaul who wrote (3743)	11/26/1998 11:25:00 PM
From: Spots	Respond to of 14778

Watts vs Volt-Amps ZP, somehow I think you got more than you bargained for <gg>. So, how about some more? In a purely resistive circuit, watts (power) is equal to volts x amps. However, introduce inductance (coils, transformers) and capacitance (well, capacitors), and the circuit has other forms of impedence which, though measured in ohms like resistance, do not in themselves draw net power ( see below). Nevertheless, inductive and capacitive impedences, do place higher voltage and current loads on the source of power. In a sense, even though the power supply will get it back, it has to supply it up front. The ratio of Watts/VA is called the power factor, which in a purely resistive circuit is 1. That's your toaster. Add transformers (inductances) and ripple filters (capacitors) and etc, and you add non-resistive impedences that will reduce the power factor. Public power companies are required to deliver AC power that is very close to a power factor of 1 (within a fraction of a percent). I don't know what the power factor of a good PC power supply is, but I would expect that it is also pretty close to one (the cheaper the lower, probably). Apparently the switching UPSs run a net capacitive load which reduces the power factor to approximately .7. I'm no electrical engineer, but I can see that it would be a lot harder to filter and smooth essentially square or stepped waves from a battery without introducing net non-resistive impedence, because these impedences are frequency sensitive. A square wave (eg) has lots of different sine wave frequencies in it (all the odd ones, actually), so the impedence of the ciruit would net out much different than if you fed a single-frequency sine wave into it. Anyhow, the fairly standard .7 power factor for UPSs has to be applied to figure the capacity. Sorry, again went on way too long. Spots () Technical aside -- these impedences do take energy (power) from the circuit during parts of the AC cycle, store it temporarily in electric (capacitance) or magnetic (inductance) fields then give it back during other parts of the cycle.